EDIT: this question of mine has received little attention, perhaps in part because it was stated in a too general and complicated way. So let me give it a second chance:

Fix an integer $r \geq 0$. Let $E_r(x)$ be the number of square free integers less than $x$ having exactly $r$ prime factors $p$ congruent to $2 \pmod{3}$ (and an arbitrary number of other prime factors). What is the order of magnitude of $E_r(x)$ when $x$ goes to $\infty$ ?

When $r=0$, one has $E_0(x) \asymp x/\log^{1/2}(x)$ (see below). Even when $r=1$,
I don't know the answer, or a lower bound for $E_1(x)$ that is better than this one for $E_0(x)$. Thanks for any
idea or reference pointing to a method that can be used to solve this question.

ORIGINAL POST FOLLOWS:

I begin by quoting a theorem found in Serre's paper "divisibilité de certaines functions
arithmétiques", with a proof attributed to Raikov, Winter, Delange (theorème 2.4 - I have slightly changed and simplified the statement, as the $E$ here is the complement $E'$ of $E$ in Serre's paper.).

Let $E$ be a set of integers satisfying the following condition :

(MM) For $n,m$ relatively prime integers, $nm \in E$ if and only if $n \in E$ and $m \in E$,

Let $P$ be the set of primes that are in $E$. Assume that $P$ has analytic density $\alpha > 0$. Let $E(x)$ the number of elements of $E$ less than $x$. Then when $x$ goes to infinity,
$$E(x) \sim c x /\log^\{1-\alpha}(x)$$ for some constant $c>0$.

Example: Let $P$ be a set of primes of density $\alpha>0$, $E$ the set of square-free
integers whose prime factors are all in $P$. Then $E$ satisfies the hypothesis (MM).
So does $F=$ the set of all integer who's prime factors are all in $P$. Both satisfies
the conclusion of the theorem, though with a different constant $c$.

Now let $P,E$ be as above, and let $r \geq 0$ be an integer. Let $E_r$ be the set of integers that are products of an element of $E$ and exactly $r$ distinct prime not in $P$. My question is:

What is an order of magnitude of $E_r(x)$ ?

The case $r=0$ is the theorem above, and I suppose the answer for $r>0$
does only depend on $P$, not on $E$, as for $r=0$. But in case it is not true, or in case that helps, the case of interest for me is the case where $E$ is as in the example.
So if I may reformulate my question in this special case: if $P$ is a set of prime of density $\alpha>0$, and $E$ is the set of square-free integers which are product
of $r$ primes not in $P$ and arbitrary number of primes in $P$, what is the order of magnitude of $E_r(x)$?

A word on my motivation, which turns around the beautiful aforementioned paper of Serre.
Consider for example the modular form $\Delta$, modulo $3$. if $E$ is the set of $n$ such that the $n$-th coefficient of that form is non-zero, that is such that $3 \nmid \tau(n)$, then $E$ satisfies (MM), $P$ is the set of primes congruent to $1$ modulo $3$, and $E(x) \sim cx log^{1/2}(x)$.
This is in Serre, as $\Delta$ is an eigenform. Noe consider $\Delta^2$ modulo $3$ which is not an eigenform, and let $F$ the set of $n$-th coefficient of that form is non-zero. Then it is easy to see that $F$ contains $E_1$ (and I think, is not too much bigger than $E_1$).

Thanks a lot : this is what I expected but was not able to prove.
–
JoëlApr 22 '12 at 20:04

You're most welcome. For the general case, one seems to need estimates for $\sum_{\substack{p \leq x\\p \in \mathbb{P}\setminus P}}\frac{1}{p}$. Perhaps that involves some generalizations of Mertens' theorem?
–
user22202Apr 23 '12 at 0:16

This is a sketch, but I would think the details are reasonably routine:

First you compute the density of the set $F_r$ of numbers having exactly $r$ prime factors equal to $2$ mod $3$ and NO other factors. The computation should be completely parallel to the computation of density of $r$-almost primes, which is discussed at length, for example, in this question. (or google for more references). call the resulting density $\delta(r);$ for primes the density is $1/\log x$

Since you already know the result for $r=0$ ($f(x) = x/(\log^{1/2} x)$), you simply compute the integral
$\int_1^N \lfloor f(N/x)\rfloor d \delta(x).$

(it seems clear that the result for $k$-almost primes in at least the original Landau form can be derived in exactly this fashion).